1. Technical Field
The present disclosure relates to a mobile storage device, in particular, to a mobile storage device that can increase heat dissipation capability.
2. Description of Related Art
A mobile storage device (e.g. a USB flash disk) generally possesses a Universal Serial Bus (USB). When the mobile storage device is coupled to a host device, the mobile storage device is operated by the power supplied by the USB and the mobile storage device exchanges data with the host device via the USB. The current USB mainstreams are USB 2.0 and USB 3.0, which are the second version and the third version of the USB respectively.
An operating voltage used for the USB is 5V (volt). However, operating voltages for a sub-circuit and an integrated circuit (IC), such as the core IC or the flash memory component, of the circuitry in the mobile storage device are usually lower than 5V. Hence, a buck circuit is usually utilized for lowering the voltage from 5V to, for instance, 1.2V or 3.3V. The buck circuit can utilize a Low DropOut linear regulator (LDO) or a pulse width modulation (PWM) circuit, wherein cost of the LDO is relatively cheaper.
Please refer to FIG. 1. FIG. 1 is a circuit block diagram of a buck circuit of a conventional mobile storage device. As shown in FIG. 1, a LDO 101 lowers a voltage of 5V to 3.3V. A LDO 102 further lowers the voltage of 3.3V to 1.2V. The voltage of 1.2V can then be supplied to a core IC (i.e. the operating voltage can also be 1.8V or 1.0V, etc., according to implementation variation of the core IC). A LDO 103 lowers the voltage of 5V to 3.3V, and the voltage of 3.3V can be supplied to a flash memory component or an input/output (I/O) port in the circuitry of the mobile storage device. Power loss P for each of the LDOs 101, 102 and 103 can be represented by a formula of P=ΔV*I, wherein AV represents a voltage difference generated by the LDO and I represents a current flowing through the LDO. For instance, when the current flowing through the LDOs 101 and 102 is 150 mA (milliamp), the total power loss for the LDOs 101 and 102 is (5V−1.2V)*150 mA=570 mW (milliwatt). When the current flowing through the LDO 103 is 100 mA (milliamp), the total power loss for the LDO 103 is (5V−3.3V)*100 mA=170 mW (milliwatt). Power loss (it will be converted to heat) of a LDO increases as a voltage drop of the LDO increases, and consequently the LDO generates more heat.